# polynomials with similar maxima-minima

Assume $p$ and $q$ are n-variate degree d, homogeneous polynomials. Define $|p|_{\infty}= \max_{x \in S^{n-1}} |p(x)|$ $D(p)=\{ x \in S^{n-1} : p(x)= |p|_{\infty} \}$
$E(p)=\{x \in S^{n-1} : p(x)= -|p|_{\infty} \}$

Define $D(q)$ and $E(q)$ similarly. Now assume $D(p) \cup E(p) = D(q) \cup E(q)$ What kind of relation we can deduce between $p$ and $q$ by this assumption?

• A polynomial of even order can be bounded from one direction, but no nonconstant polynomial in the whole space is bounded. It would make a reasonable question to ask what is the relation of two such polynomials that reach their extremum in the same set. Would something like this work for you? Mar 17 '15 at 6:50
• I've changed the question to make more sense :) This version would be helpful for me.
– alpx
Mar 18 '15 at 0:00
• Do you happen to know of any examples where $p$ and $q$ are not in a linear relation? Mar 18 '15 at 8:45
• An answer to my question: if any $r$ has degree $e$ then $r^3$ and $r\big(\sum{x_i^2}\big)^e$ would have the same extreme sets, without being linearly related (in either $\mathbb{R}^n$ or $S^{n-1}$). Mar 18 '15 at 12:37

I think that some computing around shows that $(x^6+y^6)^5$ and $(x^{10}+y^{10})^3$, or $x^{12}+y^{12}$ and $(x^4+y^4)^3$ for that matter, have the same set of extremes. So there can't be an algebraic relation between $p$ and $q$ holding in general.
• @alpx $(x^{2m}+y^{2m})/(x^2+y^2)^m$ has absolute max points at $(\pm 1, 0), (0, \pm 1)$ and absolute min points at $(\pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{2}}{2})$. These are then max/min points on $S^1$ too since $x^2+y^2$ is 1 there. Raising to odd powers then only changes the values, not the locations, of extreme points. I think similar result hold in higher dimensions too. Mar 19 '15 at 16:41